= Polynomial projections for a coprime decomposition
{title2=$P=b(T)q(T),\quad Q=a(T)p(T)$}
Suppose $p(T)q(T)=0$ for a <linear operator> $T$, and <Bézout's identity for polynomials> gives $ap+bq=1$. The <kernel decomposition for coprime polynomials> is $V=\ker p(T)\oplus\ker q(T)$. The displayed operators are the corresponding <projection operators>: $P$ is the identity on $\ker p(T)$ and zero on $\ker q(T)$, while $Q$ has the opposite action. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$ and $PQ=QP=0$. These actions also prove independence from the choice of Bézout coefficients. Every summand is an <invariant subspace> of $T$, since the projections are <polynomials> in $T$ and commute with it.
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